Variational principle
Statement 1:
The complete set of exact eigenfunctions of H define an
orthogonal complete basis set for the total space of wave
functions.
demonstration:
H is Hermitic. Let be Ya and Yb two normalized wavefunctions (associated with two
different values Ea and Eb ). We have therefore (due to hermiticity)
< Ya |H| Yb > = Ea< Ya | Yb >
< Ya |H| Yb > = Eb < Ya | Yb >
Wherefrom (Ea - Eb) < Ya | Yb > = 0
and since Ea Eb
< Ya | Yb > = 0
Consequence:
If so, it is possible to express any function as a linear combination
of the exact eigenfunctions, Yi.
1
Variational principle
Statement 2:
The energy associated with Y a function is always above that of
the eigenfunction of lowest energy: E0.
Y is not a eigenfunction; it is associated with an energy <E> that is an average energy
(mean value)
A mean value is always intermediate relative to extreme:
Greater than the smallest!
<E> > E0
2
Mean value
• If Y1 and Y2 are associated with the same
eigenvalue o: O(aY1 +bY2)=o(aY1 +bY2)
• If not O(aY1 +bY2)=o1(aY1 )+o2(bY2)
we define <O> = (a2o1+b2o2)/(a2+b2)
Dirac notations
3
Variational principle
Statement 2:
The energy associated with Y a function is always above that of
the eigenfunction of lowest energy: E0.
Ya and Yb are two eigenfunctions
< Ya |H| = < Ya | Ea and |H| Yb > = Eb | Yb>
wherefrom < Ya |H| Yb > = Ea dab
From statement 1, Y is a linear combination of Ya s
| Ya >= Sa ca| Ya >
let multiply the left by < Ya |; this leads to only one term ca= < Ya |Y >
and similarly c*a= < Y |Ya >
then
| Y >= Sa < Ya |Y > | Ya >
4
Variational principle
Statement 2, normalization:
Ya and Yb are two eigenfunctions
associated to Ea and Eb
From statement 1, Y is a linear combination of Ya s
| Y >= Sa ca| Ya >
then
< Y | Y > =Sa,b ca < Ya |Yb> cb
< Y | Y > = Sa, ca 2 < Ya |Ya> = Sa, ca 2 = 1
5
Variational principle
Statement 2, Demonstration:
< Y |H | Y > = Sa,b ca < Ya |H | Yb > c b
< Y |H | Y > = Sa,b Ea < Ya | Yb > c b
< Y |H | Y > = Sa,b Ea ca dab c b
< Y |H | Y > = Sa,Ea ca 2
E > E0
> Sa,E0 ca
2
= E0
An non-exact solution has always a higher energy than the lowest exact solution
6
Using Dirac notation
Ya and Yb are two eigenfunctions < Ya |H| = < Ya | Ea and |H| Yb > = Eb | Yb >
wherefrom < Ya |H| Yb > = Ea da
From statement 1, Y is a linear combination of Ya s
| Ya >= Sa ca| Ya >
Let multiplie the left by < Ya |; this leads to only one term ca= < Ya |Y >
and similarly c*a= < Y |Ya >
then < Y | Y > = Sa,b ca < Ya |Yb> cb
< Y | Y > = Sa,b < Ya |Y > < Ya |Yb> < Yb|Y > = Sa,b < Ya |Y > dab < Yb|Y >
< Y | Y > = Sa, |< Ya |Y >| 2 =1
Demonstration < Y |H | Y > = Sa,b < Y | Ya > < Ya |H | Yb > < Y b |Y >
< Y |H | Y > = Sa,b Ea < Y | Ya > < Ya | Yb > < Y b |Y >
< Y |H | Y > = Sa,b Ea < Y | Ya > dab < Y b |Y >
< Y |H | Y > = Sa,Ea |< Ya |Y >| 2
< Y |H | Y >
>
> Sa,E0|< Ya |Y >| 2 = E0
E0
7
Variation principle
Given an approximate expression that depends on
parameters, we must determine the parameters
to minimize the energy.
Within LCAO, an MO is a linear combination of
AOs: We have then to chose the coefficients so
to minimize the energy.
Note that the variational principle is not restricted to cases
of linear combinations.
8
A mean value is always higher than the lowest value !
c 32
c 22
c 12
c 02
9
Slater exponent for He
F=
where Z* is a parameter
<E> = - Z*2/2 u.a.
<T> = + Z*2/2 u.a.
<V> = <F(1,Z*)|-Z*/r|F(1,Z*)>
using <F(1,Z*)|1/r|F(1,Z*)> = Z* u.a.
Next slide
<V> = -Z*<F(1,Z*)|1/r|F(1,Z*)> = -Z*2u.a.
10
<F(1,Z*)|1/r|F(1,Z*)> = Z* u.a.
demonstration
11
Slater exponent for He
F=
where Z* is a parameter
12
Slater exponent for He
F=
where Z* is a parameter
Search for the energy minimum
13
LCAO
Y is a linear combination of N fa s
| Y >= Sa ca| fa >
Atomic orbitals
an approximate function
Worse description than Y
Using parameters: ca
There are N coeffcients to determine
Let minimize E with respect to each of them
Therefore, there is a set of N equations, dE/dc = 0
This seems fine; however there is a problem; Which is the problem?
14
Lagrangian
There is a constraint due to normalization: the
N coefficients are not independent.
We search the minimum of a function L of N+1 variables
Which “makes no difference with H”
L = <Y | Y > - E [<Y | Y > - 1 ]
There is a set of linear expressions to derive
dL/dc = 0
Joseph Louis de Lagrange
French
1736 - 1831
Should be 0
15
Secular determinant
Solving | Hij-ESij| =0, we find the Eis.
This is a N degree equation of E to be solved.
From N AOs, we find N energies and N MOs.
We find simultaneously E and Y (eigenfunction and eigenvalues).
16
LCAO
Assuming no change in the AOs *, the MOs
correspond to a unitary change of the basis set
of AOs.
MOs are orthogonal and normalized.
Sum over r atoms
for a given i orbital
Neglecting overlaps, Src2ir=1 for a MO, Yi
and Sic2ir=1 for a AO, fi ; the AOs are distributed
among the MOs
Sum over i orbitals for a given r atom
* This is not the case for self consistent methods.
17
Schrödinger equation for LCAO
HY = EY
Erwin Rudolf
Josef Alexander
Schrödinger
Austrian
1887 –1961
H SiciFi = E SiciFi
Let multiply on the left by Fj, integrate and
develop
Sici< |H | Fi > = E Sici < Fj |Fi >
Sici (Hij-E Sij ) = 0
This is the set of linear equation solved for
| Hij-E S ij |= 0 secular determinant
The complete set of MOs are solution (not only the lowest).
Our first approach from the variational principle emphasized the solution of lowest energy
Every extreme (derivative=0) is a physical solution!
18
Hückel Theory
Erich Armand Arthur Joseph
Hückel
German 1896-1980
Never awarded the Nobel
prize!
Also known for DebyeHückel theory
of electrolytic solutions
In the 1930's a theory was devised by
Hückel to treat the p electrons of
conjugated systems such as aromatic
hydrocarbon systems, benzene and
naphthalene.
Only p electron MO's are included
because these determine the general
properties of these molecules and the
s electrons are ignored. This is
referred to as sigma-pi separability.
The
extended
Hückel
Theory
introduced
by
Lipscomb
and
Hoffmann (1962) applies to all the
electrons.
19
Hückel Theory
We consider only 2pZ orbitals
Two parameters:
The atomic 2p energy level:
E(2pZ) = a (~ -11.4 eV)
The resonance integral for adjacent
2p orbitals
Hij(C=C) = b (~ -3 eV)
Erich Armand Arthur Joseph
Hückel
German 1896-1980
Never awarded the Nobel
prize!
Also known for DebyeHückel theory
of electrolytic solutions
E = a + x b  -x = (a – E)/ b
We chose defining b as negative
Sij = dij (the overlap is neglected).
20
SLATER KOSTER
Slater Koster for metal atoms : 9 orbitals s, p et d
Several overlaps
ss, sps, sds,
pps, ppp, pds, pdp,
dds, ddp, ddd.
Phys. Rev B 94 (1954) 1498
21
EHT Theory
• Only valence orbitals are described
• AOs are Slater orbitals
– Parameters are the Hii (atomic energy levels)
– The Slater exponents
• The Overlap, Sij, are rigorously calculated
from the geometry and the AOs
• The resonance integrals, Hij, are derived
from the Sij. Wolfsberg-Helmoltz formula
22
Hij = 1.75 (Hii+Hjj)/2 Sij
Polyenes For linear and cyclic systems (with n atoms), general solutions
exist.
Linear:
Cyclic:
For linear polyenes the energy gap is given as:
Charles Alfred Coulson
(1910-1974), English
23
Extension to heteroatoms
parameter
Coulombic
Integral
Resonance
Integral
Carbon
aC = a
bCC = b
Oxygen 1 p
electron
aO = a + b
bCO = b
parameter
Coulombic
Integral
Resonance
Integral
Sulfur
with 2 p
electrons
aS = a +
0,5b
bCS = 0,4b
aS = a +
0,2b
bCS = 0,6b
Oxygen 2 p aO = a +
electrons
2b
bCO = 0,8b
Sulfur
with 1 p
electron
Nitrogen 1
p electron
aN = a +
0,5b
bCN = b
Fluorine
aF = a +
3b
bCF = 0,7b
Nitrogen 2
p electrons
aN = a +
1,5b
bCN = 0,8b
Chlorine
aCl = a +
2b
bCCl = 0,4b
Sulfur 2 p
electrons
aS = a +
0,5b
bCS = 0,4b
Bromine
aBr = a +
1,5b
bCBr = 0,3b
Sulfur 1 p
electron
aS = a +
0,2b
bCS = 0,6b
CH3 group
hyperconju
gation
aMe = a +
2b
bCMe = 0,7b
24
| Hij-E Sij |= 0
f1
f2
f3
f4
f5
f6 …
f1
-x
1 or 0
1 or 0
1 or 0
1 or 0
1 or 0
f2
1 or 0
-x
1 or 0
1 or 0
1 or 0
1 or 0
f3
1 or 0
1 or 0
-x
1 or 0
1 or 0
1 or 0
f4
1 or 0
1 or 0
1 or 0
-x
1 or 0
1 or 0
f5
1 or 0
1 or 0
1 or 0
1 or 0
-x
1 or 0
f6
:
:
1 or 0
1 or 0
1 or 0
1 or 0
1 or 0
-x
1 means that the atoms 2 and 4 are
connected, The determinant is
symmetric relative to the diagonal
=0
25
| Hij-E Sij |= 0
f1
f2
f3
f4
f5
f6 …
f1
-x
1 or 0
1 or 0
1 or 0
1 or 0
1 or 0
f2
1 or 0
-x
1 or 0
1 or 0
1 or 0
1 or 0
f3
1 or 0
1 or 0
-x
1 or 0
1 or 0
1 or 0
f4
1 or 0
1 or 0
1 or 0
-x
1 or 0
1 or 0
f5
1 or 0
1 or 0
1 or 0
1 or 0
-x
1 or 0
f6
:
:
1 or 0
1 or 0
1 or 0
1 or 0
1 or 0
-x
=0
This is an N degree equation of x
26
| Ei-x |= 0
solving is a diagonalization problem
y4
y6 …
y1
y2
y3
y5
y1
E1-x
0
0
0
0
0
y2
0
E2-x
0
0
0
0
y3
0
0
E3-x
0
0
0
y4
0
0
0
E4-x
0
0
y5
0
0
0
0
E5-x
0
y6
:
:
0
0
0
0
0
E6-x
=0
A set of first degree equations: E=Ei associated with Yi
27
The diagonalization is a unitary transformation
OMs are orthogonal thus Srcricrj = dij
If i=j, the sum of the coefficient’s squares for a given atom is 1
(normalization)
If i≠j, the sum of the products of coefficients is 0 (orthogonality)
The matrix of coefficients is unitary
wherefrom = Sicricsj = dij
If i=j, the sum of the coefficient’s squares for a given orbital is 1
AOs are distributed among all the MOs
If i≠j, the sum of the products of coefficients is 0
If all the orbitals were filled, there would be no interaction:
all the r-s bond indices should be zero.
28
C2 H 4
2 e in 2 orbitals as in H2
We have solved the problem using symmetry
and without solving the secular equation.
f1
f2
f1
-x
1
f2
1
-x
=0
Inputs are what is in the secular determinant : a b and connectivity
Outputs are orbitals, energies, total energies, charges and bond indices
29
C2H4
without overlap
f1
f2
f1
-x
1
f2
1
-x
=0
Pu Bonding
Pg antibonding
Energy
x=1
x=-1
coefficients
-c1 + c2 = 0
c1 = c 2
c1 + c 2 = 0
c1 = -c2
0
0
-x c1 + c2 = 0
C=1/√2
Charge on atoms:
1 – Si ni cri2
Bonding
lrs= Si
ni cricsi
N-1 linear equations; the last one is redundant.
1 for pu2
0 for pupg
-1 for pu2
30
C2H4
including overlap 4e repulsion
E2 = (b -ES)2  E = ± (b -ES)
f2
f1
-E
b -ES
f2
b -ES
-E
=0
Pu Bonding
Pg antibonding
Energy
b/ (1+S)
-b/ (1-S)
coefficients
c1 = c 2 = c
c = 1 /√(2+2S)
0
c1 = -c2 = c
c = 1 /√(2+2S)
0
Mulliken
1
f1
Charge on atoms:
– Si ni (cri2 –Ss
cri csi )
Bonding
OPrs= Si
ni cricsiSrs
Overlap populations
S/(1+S)
for pu2
S/(2+2S)- S/(2-2S)
= -4S2/(4-4S2) ~ - S2
for pupg
31
2 OA interaction
modifying a
-b2/
D
-D/2
D/2
b 2/ D
The geometric mean
of b and D/2
f1
f2
f1
D/2 -E
b
f2
b
-D/2 -E
=0
(D/2 -E) (-D/2 -E) = b2  E2 - (D2 /4) = b2
E2 = b2 + (D2 /4)
E = ±√[b2 + (D2 /4)]
•If D = 0, E+ = b
•If b <<D,
E+ = (D/2) (1 + 4 b2/ D2) 0.5
E+ = (D/2) (1 + 2 b2/ D2) = D/2 (1 + 2 b2/ D)
E+ - E = b2/ D
 2nd order Perturbation term
32
C2
C1
C4
Butadiene
C3
C1
C2
C3
C4
The topology is C1 bond C2, C2 bond to C3 and C3 bond to C4
A linear model contains the information with more symmetry (the topology
does not distinguish between cis and trans)
f1
f2
f3
f4
f1
-x
1
0
0
f2
1
-x
1
0
f3
0
1
-x
1
f4
0
0
1
-x
=0
33
Conservation of Orbital
Symmetry
H C Longuet-Higgins E W Abrahamson
The Molecular orbitals are solution of the symmetry operators of the molecule.
MOs from different symmetry groups do not mix.
Hugh Christopher Longuet-Higgins
1923-2004
34
Butadiene
S or A
using symmetry for the
topology
C1
S
0
0
A
C2
C3
C4
35
Butadiene
S
Symmetric
C1
=0
C2
C3
C4
Mind that symmetry 1
is for all the atoms
Not the reduced part
x2 - x - 1 = 0
x = (1 ± √5)/2 Golden numbers 1.618 and -0.618
Coefficients - (1 ± √5)/2 c1 + c2 = 0 and normalization c12 + c22 = ½
c = 0.3717 and 0.6015
1.618
-0.618
36
olden ratio 
Golden ratio conjugate
F
37
Golden ratio
Golden ratio conjugate
a= b(1+√5)
a= √(b2+(2b)2) =b(1+√5)
b
2b
38
Polyclete and Durer
39
x2 = 1+x
The medial right triangle of this "golden" pyramid (see diagram), with
sides
is interesting in its own right, demonstrating via the
Pythagorean theorem the relationship
or
40
Fibonacci recursion,
irrational numbers (incommensurable)
Rabbit population, assuming that:
In the "zeroth" month, there is one pair of
rabbit
In the first month, the first pair begets another
pair
In the second month, both pairs of rabbits
have another pair, and the first pair dies.
In the third month, the second pair and the
new two pairs have a total of three new pairs,
41
and the older second pair.
Spirale
constructions,
in nature,
in music..
Related ?
The Doctrine of the Mean (中庸, py
Zhōngyōng) is one of the Four
Books, part of the Confucian
canonical scriptures.
42
Pentagon, icosahedra
 =1.618
F=1/ =0.618
2
43
Butadiene
Symmetric
S
C1
C2
C3
C4
Recipe to build the same determinant
Express along the AOs and gather interactions with equivalent atoms
Mind that symmetry reduction makes the normalization incomplete.
44
Butadiene
Antisymmetric
A
C1
C2
C3
C4
Mind that symmetry 1
is for all the atoms
Not the reduced part
x2 + x - 1 = 0
x = - (1 ± √5)/2 Golden numbers 0.618 and -1.618
Coefficients (1 ± √5)/2 c1 + c2 = 0 and normalization c12 + c22 = ½
c = 0.3717 and 0.6015
0.618
-1.618
45
Butadiene
Antisymmetric
A
C1
C2
C3
C4
Recipe to build the same determinant
Express along the AOs and gather interactions with equivalent atoms
Mind that symmetry reduction makes the normalization incomplete.
46
Butadiene
Small 0.3717
Large 0.6015
Large 0.6015
Small 0.3717
Large 0.6015
Small 0.3717
The number of nodes increases, the amplitude is large at
the middle for the extreme; it is large on the edges for the
Frontier orbitals.
47
Butadiene
-.3717
-.6015
-1.618
0.3717
0.6015
0.6015
-.3717
-.618
0.6015
-.3717
-.6015
0.3717
0.618
-.3717
0.6015
0.6015
1.618
0.6015
0.3717
48
Butadiene
Bonding orders Ground state
0.894
0.894
0.447
No initial statement that distinguishes 1-2 from 2-3
49
Butadiene
Bonding orders Ground state
1.35Ả
C=C – C=C
1.35Ả
1.46Ả
No initial statement that distinguishes 1-2 from 2-3
50
Matches the Lewis formula and indicates the stabilization due to delocalization
Butadiene
Bonding orders Excited state
C–C=C–C
0.447
0.447
0.724
51
It is no use to iterate
The Hückel method is powerful since predictive.
It is possible to make the parameterization
dependent on the differences in bond lengths
choosing b12 = b34 larger than b23 or modifying a
according to the charge.
This does not make the calculation predictive.
It is possible to make iterations (w technique) to
make the results “self consistent”; this is not
recommended. This does not allow controlling
the parameters.
52
Annulenes
f1
f2
f3
f1
-x
1
0
f2
1
-x
f3
0
...
...
fN-1
fN-1
0
0
1
1
0
0
0
1
-x
1
0
0
0
0
1
-x
1
0
fN-1
0
0
0
1
-x
1
fN
1
0
0
0
1
-x
=0
Secular equation:
b cr-1 -E cr + b cr+1 = 0
53
Delocalization
54
Annulenes
solutions for CN
Rq (fr) = fr+s
An annulene is a periodic system;
angle q = 2ps/N
The new coefficient for r
is the old one multiplied
by a constant
Solutions for annulenes are those of any periodic system: crystals (Bloch sums)
55
annulenes
The system allows solutions Rjq
with angle (qj=2jp/N).
There is N rotations possible including identity (j=0 ou N) : the
« simple » rotation q and the « multiple» rotations jq
Functions satisfying to rotations jq=2jp/N:
We build linear combinations of AOs,(Scrfr) satisfying
rotation qj
(Scrfr) = e-ijqj (Scrfr) with cr (j) = N-1/2 eijrqj = e –i2pjr/N .
There are N solutions, with a quantum number: j.
.
i is the square root of -1; i2=-1
56
Rotation by 120° = 2p 2/6
(j=2, orbital y2)
N exp(i2p 4/6)
r=2
N
r=3
N
r=1
C5 x exp(-ijqj)=
N exp(i2p 4/6) N exp(i2p 10/6) x
exp(-i2p 2/6) =
r=0
exp(i2p 2/6)
r=1
r=5
N exp(i2p 2/6)
0
r=4
r=0
N
N exp(i2p 8/6)
0
r=2
N exp(i2p 8/6)
r=5
r=4
N
r=3
N exp(i2p 10/6)
N exp(i2p 10/6)
The 5th AO takes the place of the first
C5 x exp(i2p 2/6)
57
annulenes
They are the LCAO functions satisfying the rotations qj
Yj= Scrfr
There are j solutions; the eigenvalues are e-ijsq
58
Benzene
annulenes
Two real solutions and
4 complexes
Degenerate by pairs
+
+
The complexe solutions are associated with rotations only; they are not
eigenfunctions of the other symmetry operators and must be the degenerate
59
annulenes
Real solutions
Satisfying two
mirror symmetry
j = 3, AS
j = ± 2, AA
j = ± 2, SS
j ± 1, SA
j = ± 1, AS
j = 0, SS
60
annulenes
Orbitals from
symmetry only
j = 3, AS
j = ± 2, AA
All the
coefficients
are equal
c=1/2
All the
coefficients
are equal
c=1/√6
j = ± 2, SS
E = 4*(< 1/2f2IHI1/2f3>+…)
=b
j = ± 1, AS
j ± 1, SA
j = 0, SS
E=< 1/√6f1+.. .IHI1/√6f1+…>
= 12*(< 1/√6f1fIHI1/√6f2>+…) =
61
2b
annulenes
Orbitals from
symmetry only
j = 3, AS
The total density in E
orbital must respect
symmetry
j = ± 2, AA
j = ± 2, SS
Degenerate orbitals:
energy must be b
j = ± 1, AS
j ± 1, SA
1/22+c2=2/6 → c= 1/√12
02+c2=2/6
→ c= 1/√3
j = 0, SS
Normalization is
satisfied
4*1/12+2*1/3=1
62
For any annulene of large size …
the number of OMs remains in the gap
from 2b to -2b
-2 b
-b
2b
b
2b
j=3
j=±2
j=±1
j=0
63
annulenes
Chosing another
set of planes;
rotation by 60°
j = 3, AS
Combining orbitals
YS’A’ = ½ [YSA + √3 YAS ]
YA’S’ = ½ [YSA - √3 YAS ]
New symmetry appears
While old ones desappears
j = ± 2, AA
j = ± 2, SS
j ± 1, SA
j = ± 1, AS
j = 0, SS
64
Hückel determinant for Cyclobutadiene
Notations: x = ↔ E = a + x b
units b (b <0) , origin a
f1
f2
f3
f4
f1 -x
f2 1
f3 0
f4 1
1
-x
1
0
0
1
-x
1
1
0
1
-x
= 0
Two possibilities of using mirror symmetries
65
First set
1/2
1/2
1/2
1/2
66
Second set
1/2
1/√2
1/√2
1/2
67
The Jahn-Teller Theorem 1937
"any non-linear molecular system in a degenerate electronic state will be
unstable and will undergo distortion to form a system of lower symmetry and
lower energy thereby removing the degeneracy"
There is a symmetry reduction (a geometry distortion) when the a set of
HOMOs is not completely filled.
Edward Teller 1908-2003
Hungarian-American
68
Hermann Arthur Jahn 1907-1979 English
In an octahedral crystal field, the t2g orbitals occur at lower energy than the
eg orbitals. The t2g are directed between bond axes while the eg point along
bond axes.
4e-8e
3e-5e-9e
0-6e
2e-8e
3e-9e
The full octahedral symmetry takes place only when the t2g set is fully
occupied.
69
Cyclobutadiene: C4H4
C4H4 is unstable, it exists stabilized by asymmetric ligand (one eg orbital is stabilized).
It dimerizes easily.
Chain of Vn-benzenen+1: The HOMO gap vanishes for n=4; then there is a
rotation of the successive benzene loosing the full D5h symmetry.
V
V
V
If the rings are eclipsed, there is no gap; the rotation of ring opens a gap.
70
B5 Boron compounds
71
Jahn-Teller effects are grouped into two categories.
The first arises from incomplete shells of degenerate orbitals. It includes the
first-order Jahn-Teller effect and the pseudo Jahn-Teller effect.
The second arises from filled and empty molecular orbitals that are close in
energy and is the second-order Jahn-Teller effect.
The two categories have quite different physical bases. As a result,
geometric distortions produced by the first are quite small and normally lead
to dynamic effects only.
In favorable cases, the second-order Jahn-Teller effect produces very large
distortions, including complete dissociation of a molecule. This can occur
even when the relevant molecular orbitals are separated in energy by as
much as 4 eV.
72
The Jahn-Teller Theorem 1937
CuCl2 Cu2+ 9e
NiCl2 Ni2+ 8e
The shielding effect this has on the electrons is used to explain why the
Jahn-Teller effect is generally only important for odd number occupancy of
the eg level.
The effect of Jahn-Teller distortions is best documented for Cu(II) complexes
(with 3 electrons in the eg level) where the result is that most complexes are
found to have elongation along the z-axis.
73
Apparent exceptions to the theorem are probably examples of what has
been called the "dynamic Jahn-Teller effect". In these cases either the
time frame of the measurement does not allow the distortion to be seen
because of the molecule randomly undergoing movement or else the
distortion is so small as to be negligible.
For one of the copper complexes, the bond lengths are apparently
identical. If the X-ray structure of the sample is redone at varying
temperatures it is sometimes possible to "freeze" a molecule into a
static position showing the distortions.
74
Alternant conjugated hydrocarbons
Definition: Atoms of conjugated molecules could be divided into two
sets of atoms (starred and unstarred atoms) so that a member of
one set is formally bonded only to members of the other set. Many
compounds are alternant; they are not when their structure contains
an odd-member ring.
75
Alternant conjugated hydrocarbons
-E cr + Ss b cs = 0
if Yj = Sr* cr fr + Ss° cs fs is an eigenfunction
Y‘j = Sr* cr fr - Ss° cs fs is another one with Ej'=-Ej
•Orbital energy levels are symmetric relative to a level.
•If there is an odd number of orbital, one of them is
nonbonding
76
Alternant conjugated hydrocarbons
Non bonding orbital, SOMO
The nonbonding orbital has non-zero coefficient only at the
starred atoms, and the sum of the coefficient of the starred atoms
attached to a given unstarred atom must equal zero.
The secular equation -E cr + becomes Ss cs = 0 with E=0.
Applied on a starred atoms Ss cs = 0 it gives again that cs = 0
Applied on a unstarred atoms Sr* cr = 0 it tells that the unstarred
atoms belong to nodal planes.
It is easy to find the coefficients from there.
77
Allyle radical
The SOMO is localized on the terminal atoms
species):
Allyle anion
The charge is -1/2 on each terminal atom
H2C
CH2
Allyle cation
The charge is +1/2 on each terminal atom
a = 1/√2
a = -1/√2
78
Dewar method for estimating
aromaticity
Two bonds close the cycle
One bond makes the polyene
If the SOMO is
symmetric, it is
better to make 2
bonds:
S or A
S
S 2n+1
A 2n-1
If the SOMO is
Single antisymmetric, it is
carbon better to make only
one bond:
Polyene
radical
A
a
S
S
S
-a
4n+2 electrons aromatic
79
4n electrons antiaromatic
Dewar method for estimating
aromaticity
Two bonds close the cycle
One bond makes the polyene
-a
-b
pyrène
-b
a
b
a
-2a
-a
3a
5a
-3a
0
b
-b
-b
b
There is an
energy gain
by forming a
new cycle
80
The net charge of alternant hydrocarbons is zero.
The electron density on r is 1
| coeff2 of occupied MOs| = | coeff2 of unoccupied MOs |
wherefrom Sini cri2=1
demonstration :
Socc cr,occ2 + Snb cr,nb2 + Svac cr,vacc2 =1
2 Socc cr,occ2 + Snb cr,nb2 = 1
Sini cr,i2 = 1
81
The bond indices between atoms of the same set is zero.
(first order perturbation between atoms of the same set is zero).
OMs are orthogonal thus Srcricrj = dij
If i=j, the sum of the coefficient’s squares for a given atom is 1
(normalization)
If i≠j, the sum of the products of coefficients is 0 (orthogonality)
The matrix of coefficients is unitary
wherefrom = Sicricsj = dij
If i=j, the sum of the coefficient’s squares for a given orbital is 1
If i≠j, the sum of the products of coefficients is 0
Socc cr,occcs,occ + Snb cr,nbcs,nb + vac cr,vacccs,vacc =drs
2 Socc cr,occcs,occ + Snb cr,nbcs,nb = 0
Socc cr,occcs,occ + Snb cr,nbcs,nb = 0
Srnicricsj = 0
82
Butadiene is an alternant
-.3717
-.6015
-1.618
0.3717
0.6015
0.6015
-.3717
-.618
0.6015
-.3717
-.6015
0.3717
0.618
-.3717
0.6015
0.6015
1.618
0.6015
0.3717
83
Benzene is
alternant
j = 3, AS
E = -2 b
annulen
es
j = ± 2, AA
j = ± 2, SS
E = -1b
j ± 1, SA
j = ± 1, AS
E = 1b
j = 0, SS
E = 2b
84
Importance of the non bonding orbital
• for radical species, it represents the unpaired electron
•For anions or cations it monitors the charge
d = Sini cr,i2
d = 2 Socc cr,occ2 + 2 cr,nb2
d = (2 Socc cr,occ2 + 1 cr,nb2 ) + cr,nb2
d = 1 + cr,nb2
For cations d = 2 Socc cr,occ2
d = (2 Socc cr,occ2 + 1 cr,nb2 ) - cr,nb2
d = 1 - cr,nb2
For anions
85
Heteroatoms:CO
parameter
Coulombic
Integral
Resonance
Integral
Carbon
aC = a
bCC = b
Oxygen 1 p
electron
aO = a + b
bCO = b
Oxygen 2 p aO = a +
electrons
2b
fC
fO
fC
-x
1
fO
1
-x+1
=0
x2-x-1=0
O more electronegative
x= (1±√5)/2
Antibonding, large amplitude on C
-0.618
bCO = 0,8b
0
0.8506 -0.5257
1 p electron for CO
0.5257 0.8506
1
Bonding, large amplitude on O
C
1.618
O
86
allyle
1
2
3
1
-x
1
0
x3-2x=0
Solutions:
x=- √2
x=0
x= √2
2
1
-x
1
3
0
1
-x
Sym
1
2
1
-x
2
AntiSym
1
Sym
1/√2(f1+f3)
1/√2(f1+f3)
-x
f2
√2
2
1
-x
1
-x
f2
√2
-x
87
Allyle coefficients
x=0
-xc1+c2=0
c1-xc2+c3=0
c2-xc3=0
c2=0
c3=-c1 this orbital is found antisymm
c2=0 no new information
x=√2
-√2 c1+c2=0
c1-√2 c2+c3=0
c2= √2 c1
c3 = c1 this orbital is found symm
c2-xc3=0
1/√2 c3= 1/√2 c1 = c2
These expressions have be normalized
88
1/2 -1/√2 1/2
-√2
-1/√2
1/√2
0
1/2 1/√2 1/2
Charges are controlled by
the filling of the nonbonding
MO.
They appear on the
terminal atoms.
The sum of charges is the
global charge
√2
radical
cation
q1 1-2(1/2)2-1(1/√2)2 = 0 1-2(1/2)2
q2 1-2(1/√2)2 = 0
= 1/2
1-2(1/√2)2
=0
anion
1-2(1/2)2-2(1/√2)2
= -1/2
1-2(1/√2)2 = 0
89
1/2 -1/√2 1/2
-√2
-1/√2
1/√2
bond indices are
independent from the filling
of the nonbonding MO.
0
1/2 1/√2 1/2
√2
radical
l12=l23 2(1/2)(1/√2) = 0.707
cation
= 0.707
anion
= 0.707
90
Exercises
Find, using symmetry, the orbitals for
• the trimethylenemethane H2C=C(CH2)2
• the pentadiene*
• the cyclopentadiene*
*Explain the origin of the solutions x=±1 for the pentadiene
*Explain the origin of the golden number solutions for the
cyclopentadiene
91
C2
Trimethylene-methane
Symmetry orbitals :
F0
1/√3 (f1+ f2+f3)
C1
C3
MOs
}
1/√2 [F0 + 1/√3 (f1+ f2+f3)]
1/√2 [F0 - 1/√3 (f1+ f2+f3)]
1/√2 (f2-f3)
1/√2 (f2-f3)
√(2/3) f1+ 1√6 ( f2+f3)
√(2/3) f1+ 1√6 ( f2+f3)
Energies 0
C0
energies ± √3
92
C2
Trimethylene-methaneC1
C0
C3
1/√2 [F0 - 1/√3 (f1+ f2+f3)]
 

√(2/3) f1+ 1√6 ( f2+f3)

1/√2 [F0 + 1/√3 (f1+ f2+f3)]
1/√2 (f2-f3)
93
Cyclopentadienyl radical and ions
3
Sym
1
2
3
1
-x
1
0
2
+2
-x
1
3
0
1
-x+1
4
2
5
1
-x2(1-x)+2(1-x)+x=0 x3-x2-x+2=0
x=2 solution (x-2) (x2-x-1)
x= (1±√5)/2
The antisymmetric solutions are those from butadiene
94
Pentagon, icosahedra
- =-1.618b
F=1/ =0.618b
2b
95
Pentagon, icosahedra
-1.618b
This level is bonding,
preferably filled.
0.618b
2b
96
The anion is stable (aromatic)
The cation is not (antiaromaticJahn-Teller situation)
Thorium+4
97
Exercise: naphtalene
1) Write the Huckel determinant for
each symmetry group of the
naphatlene.
2) Solve the SS secular determinant
3) Deduce from the SS energy levels those for AS
4) What must be the levels for SA and AA solutions ?
5) Draw the non-bonding level of the pentadienyl radical
(C5 chain)?
6) Explain from there what are the levels SS and AS at
x=±1
98
naphtalene
SS
1
2
3
1
-x+1
1
0
2
+1
-x
2
3
0
1
-x+1
SS -x (1-x) 2-3(1-x)=0 (1-x)(-x2+x+3)=0
x=1 solution (-x2+x+3)=0
x= (1±√13)/2x = 2.30277 and -1.30277
AS solutions have opposite energies (alternant)
-1, x=-2.30277 and +1.30277
The SA and AA solutions are the 4 MOs from butadiene
± 0.618 and ± 1.618
The levels at ±1 are duplication of the non bonding MO from the
99
pentadienyle in-phase and out of-phase.
Exercises
• Calculate the energy for C3 and C5
– Linear or cyclic
– Positively charge and negatively charged.
•
Calculate the MO energies for the biphenyle.
Repeat the calculation assuming that the
central bond is only half of a C=C bond.
100